Re: template member specialization of template class
On 22 =DAn, 08:10, toton <abirba...@gmail.com> wrote:
Hi,
How can I specialize a template member function for a template class.
eg, I have a tuple like class
enum dir_type{
dir_x,dir_y};
template< typename C>
class pair_type{
typedef C value_type;
typedef pair_type<C> self_type;
private:
value_type x_;
value_type y_;
public:
pair_type(const value_type& x,const value_type& y) : x_(x),y_(y){}=
template<dir_type d>
value_type& get();
};
and want to call like
typedef pair_type<int> PT;
PT p(10,12);
int x = p.get<dir_x>();
now I want to specialize template<dir_type d> value_type& get() for
dir_x and dir_y, without specializing class pair_type for C.
1) is it possible ? if, then what is the syntax ?
2) if possible, can I write such specializations in the header file
directly ?
3) if not possible, in which way I can do this (like the way boost
tuple returns, only instead of get<1> etc, I want to have the enums as
the name i.e get<dir_x>()
any help is appreciated.
thanks
abir
Hi.
It would be something like
template<typename C>
template<>
pair_type<C>::value_type& get<dir_x>()
{
}
However I think it is not possible. As workaround you can create some
structure containing static method providing required functionality. I
think example is better:
enum dir_type
{
dir_x,dir_y
};
template< typename C>
class pair_type;
template<typename C, dir_type d>
struct GetImplementation
{
static inline typename pair_type<C>::value_type& get()
{
// Default implementation
}
};
template<typename C>
struct GetImplementation<C, dir_x>
{
static inline typename pair_type<C>::value_type& get()
{
// Specialized implementation
}
};
template< typename C>
class pair_type
{
public:
typedef C value_type;
typedef pair_type<C> self_type;
private:
value_type x_;
value_type y_;
public:
pair_type(const value_type& x,const value_type& y) : x_(x),y_(y){}
template<dir_type d>
value_type& get()
{
return GetImplementation<C, d>::get();
}
};